Question

Let q(t) represent the amount of a certain reactant present at time t. suppose that the rate of decrease of q(t) is proportional to q3(t). that is, q =−kq3, where k is a positive constant of proportionality. how long will it take for the reactant to be reduced to one half of its original amount? recall that, in problems of radioactive decay where the differential equation has the form q =−kq, the half-life was independent of the amount of material initially present. what happens in this case? does half-life depend on q(0), the amount initially present?

Answer 1

The rate of change is mathematically described as :

$d(q(t))/dt = -kq(t)^3$

using variable seperation bring d(q) and all functions of q to one side and dt and all functions of t to the other side. It should look like this:

$\frac{d((t))}{q^3} = -k.dt$

Now integrate on both sides.

$-0.5q(t)^-2 = -kt - C$

where C is constant of integration. Its value can be found using initial condition by simply putting  t=0 in the above equation

C =  $1/2q(0)$

After some algebraic simplification q(t) can be written as :

$q(t)^2  = \frac{q(0)^2}{2q(0)^2Kt + 1}$

at half time, q(T) = q(0)/2, where T = half time of reaction.

substitute this in the above equation:

$T = \frac{3}{2Kq(0)^2}$

This is the required solution. As can be seen half time (T) depends on K and q(0) that is the initial concentration.